Julianne H. Portfolio Problem


Geogebra Contest




Christmas Scene










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external image MagicCircle16OctagonNoCage.jpgexternal image louvre-paris.jpg

1. Problem 1 on Page 2


1. Two different points on the line y=2 are exactly 13 units from the point (7,14). Draw a picture of this situation, and then find the coorfinates of these points.

I drew a line from Point A (7,14) that intersects with y=2 which is Point D. This is one side of the triangle, that is 12 units long. From the equation, you know the length from point A to point C is 13 units, which is a second side of the triangle and the hypotenuse of the triangle. In order to find the base of the triangle, use the pythagorean theorem.

eq=a^2 +b^2=c^2
eq=a^2 +b^2=c^2

eq=c^2-b^2=a^2
eq=c^2-b^2=a^2


eq=13^2-12^2=5^2
eq=13^2-12^2=5^2


eq=169-144=sqrt{25}
eq=169-144=sqrt{25}


eq=5= segmentDC
eq=5= segmentDC


Perpendicular lines applet

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Applet 11/5


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Applet 11/12

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Applet 11/22

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Applet 12/3

applet

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Original line equation:

-3x+2y=0 or y=3/2x

Perpendicular line equation:

2x+3y=13 or y=-2/3x+13/3

1. Solve system of equation:

3/2x=-2/3x+13/3
13/6x=13/3
x=78/39 =2

2. My vector goes from point C(5,1) to the intersection point(2,3). Therefore, my vector is [-3,2].

3. To find the other vector, double the original vector.
2 x [-3,2]=[-6,4]

4. To find the reflected point, add the original point(5,1) to the vector.
(5,1)+[-3,2]=[2,3]

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Page 32 Problem #8



Spiderman Problem


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